5 Answers, 1 is accepted
0
Shinu
Top achievements
Rank 2
answered on 09 Oct 2012, 11:55 AM
Hi,
One suggestion is that you can use FormTemplate to open custom for. Here is the demo which implements the same.
aspx:
Thanks,
Shinu.
One suggestion is that you can use FormTemplate to open custom for. Here is the demo which implements the same.
aspx:
<EditFormSettings EditFormType="Template"> <FormTemplate> . . . </FormTemplate></EditFormSettings>Thanks,
Shinu.
0
Robert
Top achievements
Rank 1
answered on 09 Oct 2012, 12:15 PM
Thank you, but what im looking for is not a form inside the grid. I got an insertform on a different page id like to link to. So can i simple cause the "Add new record" click to send me to this page and not open any form, custom or otherwise, in the grid.
I could just add a link outside the grid to this page, but i like how it looks.
Also id like to change the display text for Add new record, if you know how i can do this, id really appreciate it.
Thanks
I could just add a link outside the grid to this page, but i like how it looks.
Also id like to change the display text for Add new record, if you know how i can do this, id really appreciate it.
Thanks
0
Accepted
Jayesh Goyani
Top achievements
Rank 2
answered on 09 Oct 2012, 12:25 PM
Hello,
Thanks,
Jayesh Goyani
protected void RadGrid1_ItemCommand(object sender, GridCommandEventArgs e) { if (e.CommandName == RadGrid.InitInsertCommandName) { // Your logic comes here e.Canceled = true; } }<MasterTableView .............. <CommandItemSettings AddNewRecordText="Your New text comes here" />Thanks,
Jayesh Goyani
0
Shinu
Top achievements
Rank 2
answered on 10 Oct 2012, 05:47 AM
Hi,
One suggestion is that you can access the buttons and set the PostBackUrl property.
C#:
Thanks,
Shinu.
One suggestion is that you can access the buttons and set the PostBackUrl property.
C#:
protected void RadGrid2_ItemDataBound(object sender, GridItemEventArgs e){ if (e.Item is GridCommandItem) { GridCommandItem item = (GridCommandItem)e.Item; Button btn = (Button)item.FindControl("AddNewRecordButton"); LinkButton Addbtn = (LinkButton)item.FindControl("InitInsertButton"); Addbtn.PostBackUrl = "Page.aspx"; btn.PostBackUrl = "Page.aspx"; }}Thanks,
Shinu.
0
Robert
Top achievements
Rank 1
answered on 10 Oct 2012, 08:32 AM
Exactly what i was looking for. Thank you very much